Using the binomial theorem, Find the first three terms in the expansion of (x^2+2y^3)^20
They are, of course,
20C0 (x^2)^20(2y^3)^0 + 20C1(x^2)^19(2y^3)^1 + 20C2 (x^2)^18(2y^3)^2
x^40 + 40x^38y^3 + 760x^36y^6
To expand (x^2 + 2y^3)^20 using the binomial theorem, we need to find the coefficients of the terms.
According to the binomial theorem, the general term of the expansion is given by:
C(n, k) * (x^2)^(n-k) * (2y^3)^k
Where C(n, k) represents the binomial coefficient, given by:
C(n, k) = n! / (k! * (n-k)!)
In this case, n = 20, and we need to determine the first three terms of the expansion. Let's calculate them:
Term 1: (x^2)^20 * (2y^3)^0
C(20, 0) * (x^2)^(20-0) * (2y^3)^0
= 1 * x^(2 * 20) * 2^0 * y^(3 * 0)
= x^40
Term 2: (x^2)^19 * (2y^3)^1
C(20, 1) * (x^2)^(19-1) * (2y^3)^1
= 20 * x^(2*19) * 2^1 * y^(3*1)
= 20x^38 * 2y^3
= 40x^38y^3
Term 3: (x^2)^18 * (2y^3)^2
C(20, 2) * (x^2)^(18-2) * (2y^3)^2
= 190 * x^(2*18) * 2^2 * y^(3*2)
= 190x^36 * 4y^6
= 760x^36y^6
Therefore, the first three terms in the expansion of (x^2 + 2y^3)^20 are:
Term 1: x^40
Term 2: 40x^38y^3
Term 3: 760x^36y^6
To find the first three terms in the expansion of (x^2 + 2y^3)^20 using the binomial theorem, follow these steps:
Step 1: Identify the values of a and b in the binomial expression (x^2 + 2y^3)^20.
In this case, a = x^2 and b = 2y^3.
Step 2: Determine the value of the exponent, n. In this case, n = 20.
Step 3: Write down the binomial theorem formula. It states that for any positive integer n:
(x + y)^n = C(n,0) * a^n * b^0 + C(n,1) * a^(n-1) * b^1 + C(n,2) * a^(n-2) * b^2 + ... + C(n,n-1) * a^1 * b^(n-1) + C(n,n) * a^0 * b^n
Where C(n, k) denotes the binomial coefficient, which is equal to n! / (k!(n-k)!). The exclamation mark (!) indicates factorial.
Step 4: Plug in the values into the formula and simplify to find the first three terms.
For the first term:
C(20, 0) * (x^2)^20 * (2y^3)^0 = 1 * x^(2*20) * 2^0 = x^40.
For the second term:
C(20, 1) * (x^2)^(20-1) * (2y^3)^1 = 20 * x^(2*19) * 2 * y^3 = 20x^38 * 2y^3 = 40x^38y^3.
For the third term:
C(20, 2) * (x^2)^(20-2) * (2y^3)^2 = 190 * x^(2*18) * (2^2) * y^(3*2) = 190 * x^36 * 4 * y^6 = 760x^36y^6.
Therefore, the first three terms in the expansion of (x^2 + 2y^3)^20 are:
x^40, 40x^38y^3, 760x^36y^6.